Typical bone is made up of D.
Answer:
1 album
Step-by-step explanation:
it would take along time to explain but i did the work i promise
Answer: ![\bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}](https://tex.z-dn.net/?f=%5Cbold%7Bc%3D%5Cdfrac%7B1%2B%5Csqrt%7B19%7D%7D%7B3%7D%5Capprox1.8%7D)
<u>Step-by-step explanation:</u>
There are 3 conditions that must be satisfied:
- f(x) is continuous on the given interval
- f(x) is differentiable
- f(a) = f(b)
If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.
f(x) = x³ - x² - 6x + 2 [0, 3]
1. There are no restrictions on x so the function is continuous ![\checkmark](https://tex.z-dn.net/?f=%5Ccheckmark)
2. f'(x) = 3x² - 2x - 6 so the function is differentiable ![\checkmark](https://tex.z-dn.net/?f=%5Ccheckmark)
3. f(0) = 0³ - 0² - 6(0) + 2 = 2
f(3) = 3³ - 3² - 6(3) + 2 = 2
f(0) = f(3) ![\checkmark](https://tex.z-dn.net/?f=%5Ccheckmark)
f'(x) = 3x² - 2x - 6 = 0
This is not factorable so you need to use the quadratic formula:
![x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-%28-2%29%5Cpm%20%5Csqrt%7B%28-2%29%5E2-4%283%29%28-6%29%7D%7D%7B2%283%29%7D%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D%5Cdfrac%7B2%5Cpm%20%5Csqrt%7B4%2B72%7D%7D%7B2%283%29%7D%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D%5Cdfrac%7B2%5Cpm%20%5Csqrt%7B76%7D%7D%7B2%283%29%7D%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D%5Cdfrac%7B2%5Cpm%202%5Csqrt%7B19%7D%7D%7B2%283%29%7D%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D%5Cdfrac%7B1%5Cpm%20%5Csqrt%7B19%7D%7D%7B3%7D%5C%5C%5C%5C%5C%5C.%5Cquad%20%5Capprox%5Cdfrac%7B1%2B4.4%7D%7B3%7D%5Cquad%20and%5Cquad%20%5Cdfrac%7B1-4.4%7D%7B3%7D%5C%5C%5C%5C%5C%5C.%5Cquad%20%5Capprox1.8%5Cqquad%20and%5Cquad%20-1.1)
Only one of these values (1.8) is between 0 and 3.
It is 20 ft. You divide by 20 and then multiply to check
we know that
The surface area of a cube is equal to
![S=6*A](https://tex.z-dn.net/?f=%20S%3D6%2AA%20)
where
A is the area of one face of the cube
in this problem
![A=8.5^{2}= 72.25\ in ^{2}](https://tex.z-dn.net/?f=%20A%3D8.5%5E%7B2%7D%3D%2072.25%5C%20in%20%5E%7B2%7D)
Find the surface area
![S=6*72.25=433.5\ in ^{2}](https://tex.z-dn.net/?f=%20S%3D6%2A72.25%3D433.5%5C%20in%20%5E%7B2%7D%20)
therefore
<u>the answer is</u>
of wrapping paper is needed to cover the box completely